Use spherical coordinates to evaluate the
triple integral x2 + y2 + z2 DV, where E
is the ball: x2 + y2 + z2 = 16.
1/S22+
Answers
Answer:
4096π / 5
Step-by-step explanation:
∫∫∫ (x² + y² + z²) dV
In spherical coordinates, x² + y² + z² = r², and dV = r² sin φ dr dθ dφ.
E is the range 0 ≤ r ≤ 4, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
∫₀ᵖⁱ∫₀²ᵖⁱ∫₀⁴ (r²) (r² sin φ dr dθ dφ)
∫₀ᵖⁱ∫₀²ᵖⁱ∫₀⁴ (r⁴ sin φ) dr dθ dφ
Evaluate the first integral.
∫₀ᵖⁱ∫₀²ᵖⁱ (⅕ r⁵ sin φ)|₀⁴ dθ dφ
∫₀ᵖⁱ∫₀²ᵖⁱ (¹⁰²⁴/₅ sin φ) dθ dφ
¹⁰²⁴/₅ ∫₀ᵖⁱ∫₀²ᵖⁱ (sin φ) dθ dφ
Evaluate the second integral.
¹⁰²⁴/₅ ∫₀ᵖⁱ (θ sin φ)|₀²ᵖⁱ dφ
¹⁰²⁴/₅ ∫₀ᵖⁱ (2π sin φ) dφ
²⁰⁴⁸/₅ π ∫₀ᵖⁱ sin φ dφ
Evaluate the third integral.
²⁰⁴⁸/₅ π (-cos φ)|₀ᵖⁱ
²⁰⁴⁸/₅ π (-cos π + cos 0)
²⁰⁴⁸/₅ π (1 + 1)
⁴⁰⁹⁶/₅ π
Answer:
[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \frac{4096 \pi}{5}[/tex]
General Formulas and Concepts:
Calculus
Integration
IntegralsIntegration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Multivariable Calculus
Triple Integrals
Cylindrical Coordinate Conversions:
[tex]\displaystyle x = r \cos \theta[/tex][tex]\displaystyle y = r \sin \theta[/tex][tex]\displaystyle z = z[/tex][tex]\displaystyle r^2 = x^2 + y^2[/tex][tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]Spherical Coordinate Conversions:
[tex]\displaystyle r = \rho \sin \phi[/tex][tex]\displaystyle x = \rho \sin \phi \cos \theta[/tex][tex]\displaystyle z = \rho \cos \phi[/tex][tex]\displaystyle y = \rho \sin \phi \sin \theta[/tex][tex]\displaystyle \rho = \sqrt{x^2 + y^2 + z^2}[/tex]Integral Conversion [Spherical Coordinates]:
[tex]\displaystyle \iiint_T {f( \rho, \phi, \theta )} \, dV = \iiint_T {\rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]
Step-by-step explanation:
*Note:
Recall that φ is bounded by 0 ≤ φ ≤ 0.5π from the z-axis to the x-axis.
I will not show/explain any intermediate calculus steps as there isn't enough space.
Step 1: Define
Identify given.
[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV[/tex]
[tex]\displaystyle \text{Region E (Ball):} \ x^2 + y^2 + z^2 \leq 16[/tex]
Step 2: Integrate Pt. 1
Find ρ bounds.
[Ball] Substitute in Spherical Coordinate Conversions:[tex]\displaystyle \rho^2 \leq 16[/tex]Solve:
[tex]\displaystyle \rho \leq 4[/tex]Define limits:
[tex]\displaystyle 0 \leq \rho \leq 4[/tex]
Find θ bounds.
[Ball] Substitute in z = 0:[tex]\displaystyle x^2 + y^2 \leq 16[/tex][Circle] Graph [See 2nd Attachment][Graph] Identify limits [Unit Circle]:
[tex]\displaystyle 0 \leq \theta \leq 2 \pi[/tex]
Find φ bounds.
[Circle] Substitute in Cylindrical Coordinate Conversions:[tex]\displaystyle r^2 \leq 16[/tex]Solve:
[tex]\displaystyle r \leq 4[/tex]Substitute in Spherical Coordinate Conversions:
[tex]\displaystyle \rho \sin \phi \leq 4[/tex]Solve:
[tex]\displaystyle \phi \leq \frac{\pi}{2}, \frac{3 \pi}{2}[/tex]Define limits:
[tex]\displaystyle 0 \leq \phi \leq \pi[/tex]
Step 3: Integrate Pt. 2
[Integrals] Convert [Integral Conversion - Spherical Coordinates]:[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \iiint_E {(x^2 + y^2 + z^2) \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex][dρ Integrand] Rewrite [Spherical Coordinate Conversions]:
[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \iiint_E {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex][Integrals] Substitute in region E:
[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \int\limits^{2 \pi}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^4_0 {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]
We evaluate this spherical integral by using the integration rules, properties, and methods listed above:
[tex]\displaystyle \begin{aligned}\iiint_E {x^2 + y^2 + z^2} \, dV & = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 \int\limits^4_0 {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta \\& = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 {\frac{\rho^5}{5} \sin \phi \bigg| \limits^{\rho = 4}_{\rho = 0}} \, d\phi \, d\theta \\& = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 {\frac{1024}{5} \sin \phi} \, d\phi \, d\theta\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}\iiint_E {x^2 + y^2 + z^2} \, dV & = \int\limits^{2 \pi}_0 {\frac{-1024}{5} \cos \phi \bigg| \limits^{\phi = \pi}_{\phi = 0}} \, d\theta \\& = \int\limits^{2 \pi}_0 {\frac{2048}{5}} \, d\theta \\& = \frac{2048}{5} \theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0} \\& = \frac{4096 \pi}{5}\end{aligned}[/tex]
∴ we have evaluated the given integral in spherical coordinates.
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Learn more about spherical coordinates: https://brainly.com/question/17166987
Learn more about multivariable calculus: https://brainly.com/question/4746216
---
Topic: Multivariable Calculus
Unit: Triple Integrals Applications
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Answers
Answer:
Hi,
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Any questions just ask in the comments.
Since the equation given is f(d) = 325d +8,795, the value that 8,795 represent in this situation is the initial cost.
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325d means that the cost will increase each day by that amount. Therefore, the value that 8,795 represents in this situation is the initial cost.
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[tex] \frac{1}{12} + \frac{6}{12} = \\ [/tex]
[tex] \frac{1 + 6}{12} = \frac{7}{12} \\ [/tex]
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Step-by-step explanation:
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...........
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Answers
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Refer to picture
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